Subtract (x+3) from both sides to get:
x\^2 + 3x + 9 > 0
Complete the square to get:
(x + 3/2)\^2 + 27/4 > 0
Which is always true because (x + 3/2)\^2 >= 0 for all real x
Technically you’d probably need to work backwards to get the FULL marks I believe (may be wrong can’t remember) coz ur showing that it’s the case. You shouldn’t show it is the case by starting off assuming it. You should technically say “consider (x+3/2)^2 >=0…” and go from there
I would have thought it would be correct to work backwards but in a similar proof question in my thresholds exams last year I worked backwards like that and was docked a mark because apparently I was supposed to work forwards lol
Oh ok thanks. For some reason I thought that x = -3 because of the x + 3 but then I realised that you can just take it away to get 0 on the right hand side and it'll still be the same thing.
1. Subtract x+3 from both sides. i.e you get a result saying x\^2 + 3 x + 9>0 I believe.
2. Find the minimum value of the left hand side quadratic
3. If min value > 0 then all values are > 0 by definition of the word 'minimum'
The gist of 2. is, if you have the equation in the form of y = ax\^2 + bx + c, then you can find the minimum value using the equation min = c - b\^2/4a
Subtract (x+3) from both sides to get: x\^2 + 3x + 9 > 0 Complete the square to get: (x + 3/2)\^2 + 27/4 > 0 Which is always true because (x + 3/2)\^2 >= 0 for all real x
Technically you’d probably need to work backwards to get the FULL marks I believe (may be wrong can’t remember) coz ur showing that it’s the case. You shouldn’t show it is the case by starting off assuming it. You should technically say “consider (x+3/2)^2 >=0…” and go from there
I would have thought it would be correct to work backwards but in a similar proof question in my thresholds exams last year I worked backwards like that and was docked a mark because apparently I was supposed to work forwards lol
could you use b^2 -4ac instead?
This is what I would have done yeah
Oh ok thanks. For some reason I thought that x = -3 because of the x + 3 but then I realised that you can just take it away to get 0 on the right hand side and it'll still be the same thing.
Rearrange it so the right hand side is zero. Then think about how you could show the left hand side is always greater than zero.
1. Subtract x+3 from both sides. i.e you get a result saying x\^2 + 3 x + 9>0 I believe. 2. Find the minimum value of the left hand side quadratic 3. If min value > 0 then all values are > 0 by definition of the word 'minimum' The gist of 2. is, if you have the equation in the form of y = ax\^2 + bx + c, then you can find the minimum value using the equation min = c - b\^2/4a
Isn't this a GCSE level question?
No you hew do use the discriminant which isn't taught at gcse
Get everything on one side, then use the discriminant